pub fn new_birthday_probability(n: u32) -> f64 {
    let mut answer = 1f64;
    for i in 0..n {
        // answer *= (365 - i) as f64 / 365f64;
        // 可化简如下
        answer *= 1.0 - i as f64 / 365f64;
    }
    1f64 - answer
}
